proof that all cyclic groups of the same order are isomorphic to each other

Proof.

Let GGG be a cyclic group and ggg be a generator of GGG. Define φ:ℤ→Gnormal-:φnormal-→ℤG\varphi\colon{\mathbb{Z}}\to G by φ⁢(c)=gcφcsuperscriptgc\varphi(c)=g^{c}. Since φ⁢(a+b)=ga+b=ga⁢gb=φ⁢(a)⁢φ⁢(b)φabsuperscriptgabsuperscriptgasuperscriptgbφaφb\varphi(a+b)=g^{{a+b}}=g^{a}g^{b}=\varphi(a)\varphi(b), φφ\varphi is a group homomorphism. If h∈GhGh\in G, then there exists x∈ℤxℤx\in{\mathbb{Z}} such that h=gxhsuperscriptgxh=g^{x}. Since φ⁢(x)=gx=hφxsuperscriptgxh\varphi(x)=g^{x}=h, φφ\varphi is surjective.

If GGG is finite, then let |G|=nGn|G|=n. Thus, |g|=|⟨g⟩|=|G|=nggGn|g|=|\langle g\rangle|=|G|=n. If gc=eGsuperscriptgcsubscripteGg^{c}=e_{G}, then nnndividesccc. Therefore, ker⁡φ=n⁢ℤkernelφnℤ\ker\varphi=n{\mathbb{Z}}. By the first isomorphism theorem, G≅ℤ/n⁢ℤ=ℤnGℤnℤsubscriptℤnG\cong\mathbb{Z}/n\mathbb{Z}=\mathbb{Z}_{n}.

Let HHH and KKK be cyclic groups of the same order. If HHH and KKK are infinite, then, by the above argument, H≅ℤHℤH\cong{\mathbb{Z}} and K≅ℤKℤK\cong{\mathbb{Z}}. If HHH and KKK are finite of order nnn, then, by the above argument, H≅ℤnHsubscriptℤnH\cong{\mathbb{Z}}_{n} and K≅ℤnKsubscriptℤnK\cong{\mathbb{Z}}_{n}. In any case, it follows that H≅KHKH\cong K.
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